Answer
$S^{\prime}(x)=\sec^2x(x^2-x+1)+\tan x(2x-1)$
Work Step by Step
We have: $S(x)=(x^2-x+1)\tan x$
We need to differentiate both sides with respect to $x$.
$S^{\prime}(x)=\dfrac{d}{dx} [(x^2-x+1)\tan x]\\=(x^2-x+1)\sec^2x+\tan x(2x-1)$
Therefore, $S^{\prime}(x)=\sec^2x(x^2-x+1)+\tan x(2x-1)$
